A) \[\frac{1}{n(n+1)}\]
B) \[\frac{n}{n+1}\]
C) \[\frac{2n}{n+1}\]
D) \[\frac{2}{n(n+1)}\]
Correct Answer: B
Solution :
\[\left( \frac{1}{1}-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)+.........+\left( \frac{1}{n}-\frac{1}{n+1} \right)\] \[=1-\frac{1}{n+1}=\frac{n}{n+1}\].
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